24. Balkanska matematiqka olimpijada


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1 4. Balkanska matematika olimpijada Rodos, Gka 8. apil U konveksnom etvoouglu ABCD vaжi AB = BC = CD, dijagonale AC i BD su azliite duжine i seku se u taki E. Dokazati da je AE = DE ako i samo ako je BAD+ ADC = 10. (Albanija). Na i sve funkcije f : R R takve da za sve ealne bojeve x, y vaжi f(f(x) + y) = f(f(x) y) + 4f(x)y. (Bugaska) 3. Na i sve piodne bojeve n za koje postoji pemutacija σ bojeva 1,,..., n takva da je boj σ(1) + σ() + + σ(n) acionalan. (Sbija) 4. Dat je ceo boj n 3. Neka su C 1, C i C 3 ganice ti konveksna ntougla u avni takva da je pesek svake dve od ǌih konaan skup taaka. Na i najve i mogu i boj taaka skupa C 1 C C 3. (Tuska)
2 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas, 8 Apil 007 Poblem 1. Let ABCD be a convex uadilateal with AB = BC = CD, AC BD and let E be the intesection point of its diagonals. Pove that AE = DE if and only if BAD + ADC = 10. Poblem. Find all functions f : such that ( ( ) ) ( ( ) ) ( ) f f x + y = f f x y + 4 f x y, fo any xy,. Poblem 3. Find all positive integes n such that thee is a pemutation σ of the set {1,,,n} fo which σ(1) + σ() + + σ( n) is a ational numbe. Note: A pemutation of the set {1,,,n} is a onetoone function of this set to itself. Poblem 4. Fo a given positive intege n >, let C1, C, C 3 be the boundaies of thee convex ngons in the plane such that C C, C C, C C ae finite. Find the maximum numbe of points of the set C C C 3. 1 Time allowed 4 hous and 30 minutes Each poblem is woth 10 points.
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4 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas (Apil 8, 007) Poblem. Find all functions f : R! R such that f(f(x) + y) = f(f(x) y) + 4f(x)y fo any x; y R: Solution. It is clea that the function f 0 satis es the given condition. y Assume that f 6 0: Choose x 0 such that f(x 0 ) 6= 0 and set ey = 4f(x 0 ) y: Then plugging in to the given elation x 0 fo x and ey fo y we get fo any y = f(f(x 0 ) + ey) f(f(x 0 ) ey): (1) Fo any y 1 ; y, plugging in to (1) y 1 y fo y, we get y 1 y y = f(f(x 0 ) + ^ 1 y ) f(f(x 0 ) y^ 1 y ): In othe wods, fo any y 1 ; y thee exist x 1 (y 1 ; y ) = f(x 0 ) + ]y 1 y ; x (y 1 ; y ) = f(x 0 ) + ]y 1 y such that y 1 y = f(x 1 ) f(x ), i.e. such that f(x 1 ) y 1 = f(x ) y : () On the othe hand, eplacing y by f(x) y in the given condition gives f(f(x) y) = f(y) + 4f(x)(f(x) y); i.e. f(y) y = f(f(x) y) ((f(x) y) : (3) Now if fo any two y 1 ; y, we plug in to (3), x 1 (y 1 ; y ) and x (y 1 ; y ) espectively, we get f(y 1 ) y 1 = f(f(x 1 ) y 1 ) ((f(x 1 ) y 1 ) and f(y ) y = f(f(x ) y ) ((f(x ) y ) Then by () we get f(y 1 ) y1 = f(y ) y: Since this happens fo any two y 1 ; y, we conclude that f(x) x = constant fo all x, thus f(x) = x + c; c R. It is easy to check that such a function satis es the condition of the poblem. 1
5 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas (Apil 8, 007) Poblem 3. Find all positive s integes n such that thee is a pemutation of the set f1; ; :::; ng; fo which (1) + () + ::: + p (n) is a ational numbe. Note: A pemutation of the set f1; ; :::; ng is a onetoone function of this set to itself. s Solution. Fo some n N, let (1) + () + ::: + p (n) = 1 Q. Suaing both sides of the euation we get that () + (3) + ::: + p (n) is s also ational. s Using the same easoning ecusively, we get that fo evey k f1; : : : ; ng, (k) + (k + 1) + ::: + p (n) is ational as well. Knowing that the suae oot of a positive intege is eithe intege o iational, we have that p s (n) is intege. Similaly, we get that (k) + (k + 1) + ::: + p (n), fo evey k f1; : : : ; ng, is intege. Note that s fo k = 1 we get 1 N. We de ne a k as a k = n + n + ::: + p n, fo all k 1. It is easy to {z } k pove by induction that a k < p s n + 1, fo evey k 1. Theefoe, we have (1) + () + ::: + p (n) < a n < p n + 1, implying 1 < p n + 1. Let ` be the positive intege that satis es ` n < (`+1). Fo some i; 1 i n, we have (i) = `. We distinguish two cases: Fist case: i 6= n: Then we have ` < s ` + (i + 1) + ::: + p (n) < p n+1 < `+, implying 1
6 s ` + (i + 1) + ` + 1 = ::: + p (n) = ` + 1. But then it follows that giving ` < 1. A contadiction. Second case: i = n: (i + 1) + ::: + p (n) < p n + 1 < ` + ; Fo ` > 1, ` 1 belongs to the set f(1); :::; (n 1)g. Let j < n be such that (j) = ` 1. Similaly to the st case, we have implying ` < s ` 1 + ` 1 + ` + = (j + 1) + ::: + p` < p n + 1 < ` + ; p (j + 1) + ::: + p` = ` + 1, and (j + 1) + ::: + p` < p n + 1 < ` + ; a contadiction. If ` = 1, then n f1; ; 3g. Checking though all the possibilities, it is easy to see that fo n = 1 and n = 3 thee exist pemutations that satisfy the initial condition. Namely, fo n = 1 we have p 1 = 1, and fo n = 3, we have + p 3 + p 1 =. Fo n = thee is no such pemutation.
7 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas (Apil 8, 007) Poblem 4. Fo a given positive intege n >, let C 1 ; C ; C 3 be the boundaies of thee convex ngons in the plane such that the sets C 1 \ C, C \ C 3, C 3 \ C 1 ae nite. Find the maximum numbe of points of the set C 1 \ C \ C 3. Solution 1: Let us st obseve that, if a line intesects a convex n gon at nitely many points, then the numbe of such points is at most. Theefoe any two of the n gons may intesect in at most n points. Choose two of the n gons, C 1 ; C, and say that thei intesection points ae p 1 ; p ; : : : ; p k. Thus k n. Say that the union of the set of vetices of C 1 and C is f 1 ; ; : : : ; n g. We note that it is possible to have i = j fo some i 6= j. We will de ne a onetoone function f fom fp 1 ; p ; : : : ; p k g to f 1 ; ; : : : ; n g as follows. Fist of all, oient all n gons in the clockwise diection. Thus, if one taveses an n gon accoding to this oientation, the inteio is on the ight and the exteio is on the left. Fo evey p i, thee exist pecisely two line segments (of nonzeo length) which ae subsets of C 1 o C, say [ j ; p i ] on C 1 and [ k ; p i ] on C, such that one can appoach to p i via these line segments in the clockwise diection. Suppose, that the two vectos p i j and p i k, in this ode, fom a ight handed coodinate system. Then none of the points on [ j ; p i ] can be on o in the inteio of C, since fo any point on o in the inteio of C, the vectos p i and p i k ae eithe positive multiples of each othe, o fom a left handed coodinate system. In this case we set f(p i ) = j. Othewise we set f(p i ) = k. In both cases, the agument above shows that thee ae no othe intesection points between f(p i ) and p i, in the clockwise diection. Let us now show that f is 1 1. If f(p i ) = f(p l ) = and say (without loss of geneality) belong to C 1, then the st intesection point encounteed when one stats fom and taveses C 1 in the clockwise diection has to be both p i and p l, hence p i = p l. Now let us estimate the numbe of p i s that can be contained by the thid polygon C 3. Each edge of C 3 contains exactly 0; 1 o of the p i s. Suppose that a given edge of C 3 contains of the p i s, say p 1 and p. Since C 1 and C ae convex and thei intesection with C 3 is geneic, they should have vetices between (in the clockwise sense) p 1 and p (with this ode), and outside C 3. We claim that at least one of these vetices is not in the set f(c 1 \ C \ C 3 ). Let 1 C 1 and C espectively be between (in the clockwise sense) p 1 and p (with this ode). If 1 is on o in the inteio of C (o is on o in the inteio of C 1 ), then 1 (o ) is not in the image of f, since ecall that f(p) fo any p C 1 \ C (thus fo any p C 1 \ C \ C 3 ) is a point of one of C 1 ; C not on o in the inteio of the othe. So the claim is established in this case. The emaining case is to assume that none of the vetices of any of C 1 ; C that lie outside C 3 (and between (in the clockwise sense) p 1 and p 1
8 (with this ode)) also lies in the inteio of the othe of C 1 ; C. In this case clealy the polygons C 1 and C must meet at some point between (in the clockwise sense) p 1 and p (with this ode). Say p 3 is the closest to p 1 such point. Then clealy f(p 3 ) is a vetex of one of C 1 ; C between (in the clockwise sense) p 1 and p (with this ode). These pats of C 1 ; C though, lie outside C 3 ; the inteio of C 3 lie on the othe side of the line p 1 p. Thus f(p 3 ) is not in f(c 1 \ C \ C 3 ) and the claim is established in all cases. Fo the side a of C 3 containing p 1 ; p let us call (a) a vetex as the one in the claim we just poved. It is easy to see that fo distinct sides a; b of C 3 that contain two of the p s, the points a ; b ae distinct. Indeed, let a contain p 1 ; p and b contain p 3 ; p 4 among the p s. If one of a ; b belongs to one of C 1 ; C and the othe does not belong to it, we ae okay. If both a ; b belong to say C 1, then in a clockwise tou aound C 1 stating at p 1, we meet p 1 ; p ; p 3 ; p 4 in this ode. If not, say the ode is p 1 ; p 3 ; p ; p 4. Then the segments p 1 p, p 3 ; p 4 intesect at an inteio point, since C 1 is a convex polygon. But then the sides a; b of C 3 have a common inteio point, a contadiction. So the coect ode is p 1 ; p ; p 3 ; p 4. But we know that adding (a); (b) in this tou the coect ode is p 1 ; (a); p ; p 3 ; (b); p 4. Thus (a); (b) ae distinct as claimed. Now if x of the edges of C 3 contain 1 of the p i s and y of them contain of the p i s, then x + y numbe of sides of C 3, i.e. x + y n. The numbe of points in C 1 \ C \ C 3 is x + y. Since f is injective, x + y is also the numbe of s in f(c 1 \ C \ C 3 ). Also, by the agument in the pevious paagaph, we see that fo evey distinct edge of C 3 containing points we can assign a coesponding distinct i outside the image of f(c 1 \ C \ C 3 ). Theefoe x is less o eual to the numbe of s that do not belong in f(c 1 \ C \ C 3 ). So (x + y) + y is at most as much as the numbe of s. I.e. x + 3y n. Adding this with x + y n and dividing by, and also taking into account that x + y is an intege x + y b 3n c Let us now show that this is the best uppe bound fo evey n 3. One way (among many) to constuct an example is as follows: Constuct two egula n gons C 1 ; C with the same cente, such that thei intesection points fom a egula n gon. Call the vetices p 1 ; p ; : : : ; p n in a cyclic ode. Let the cicumcicle of this n gon be C. Then let the n gon bounded by the lines p 1 p 3 ; p 5 p 7 ; p 9 p 11 ; : : : (including p k+1 p 1 in case n is an odd n = k + 1) togethe with the tangent lines to C at p 4 ; p 8 ; p 1 ; : : : be C 3. It can easily be checked that jc 1 \ C \ C 3 j = b 3nc. Solution : Let A and B be two conseutive points of C 1 \ C \ C 3 obseved in the clockwise diection fom a point in the inteio of all thee ngons. Let s look fo each C i its section in the clockwise diection between A and B exluding these points. If some two of these sections both do not contain any vetices of thei coesponding ngons, then the segment AB belongs to both ngons, a contadiction.
9 Thus at least two of these segments have at least one vetex each, and moeovethey do not contain the segment. Tivially, two distinct such vetices exist. Since thee exist jc 1 \ C \ C 3 j many conseutive points points A and B of C 1 \ C \ C 3, thee should exist at least jc 1 \ C \ C 3 j distinct vetices of the thee ngons. Thus jc 1 \ C \ C 3 j 3n i.e. jc 1 \ C \ C 3 j b 3nc since (jc 1 \ C \ C 3 j is an intege as well). Actually we can achieve this uppe bound by the example given in the Solution 1. 3
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